Question

Let X = the time between two successive arrivals at the drive-up window of a local bank. If X has an exponential distribution with λ = 1, (which is identical to a standard gamma distribution with α = 1), compute the following. (a) The expected time between two successive arrivals
(b) The standard deviation of the time between successive arrivals
(c) P(X ? 1)
(d) P(2 ? X ? 5)

Answer 1

Answer:

a) E(X)=1

b) σ=1

c) P(X>1)=0.368

d) P(2<X<5)=0.128

Step-by-step explanation:

We have X: "the time between two successive arrivals at the drive-up window of a local bank", exponentially distributed with λ = 1.

a) We have to compute the expected time between two successive arrivals.

The expected value for X as it is exponentially distributed is:

$E(X)=\dfrac{1}{\lambda}=\dfrac{1}{1}=1$

b) We have to compute the standard deviation of X.

The standard deviation is calculated as:

$\sigma=\dfrac{1}{\lambda}=\dfrac{1}{1}=1$

c) The probability that X is larger than 1 (P(X>1))

We can express the probability as:

$P(X>t)=e^{-\lambda t}\\\\P(X>1)=e^{-\lambda\cdot 1}=e^{-1}=0.368$

d) The probability that X is between 2 and 5 (P(2<X<5))

$P(X>t)=e^{-\lambda t}\\\\\\P(2<X<5)=P(X>2)-P(X>5)\\\\P(2<X<5)=e^{-\lambda\cdot 2}-e^{-\lambda\cdot 5}=e^{-2}-e^{-5}=0.135-0.007\\\\P(2<X<5)=0.128$